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Let $f\in L^1(\mathbb{R})$. Compute $\lim_{|h|\rightarrow\infty}\int_{-\infty}^\infty |f(x+h)+f(x)|dx$

If $f\in C_c(\mathbb{R})$ I got the limit to be $\int_{-\infty}^\infty |f(x)|dx$. I am not sure if this is right.

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Use density in $L^1$ of continuous functions with compact support. – Davide Giraudo Jul 11 '12 at 22:12
I tried that and I am stuck – john Jul 11 '12 at 22:18

2 Answers 2

up vote 1 down vote accepted
  • Let $f$ a continuous function with compact support, say contained in $-[R,R]$. For $h\geq 2R$, the supports of $\tau_hf$ and $f$ are disjoint (they are respectively $[-R-h,R-h]$ and $[-R,R]$ hence \begin{align*} \int_{\Bbb R}|f(x+h)+f(x)|dx&=\int_{[-R,R]}|f(x+h)+f(x)|+\int_{[-R-h,R-h]}|f(x+h)+f(x)|\\ &=\int_{[-R,R]}|f(x)|+\int_{[-R-h,R-h]}|f(x+h)|\\ &=2\int_{\Bbb R}|f(x)|dx. \end{align*}
  • If $f\in L^1$, let $\{f_n\}$ a sequence of continuous functions with compact support which converges to $f$ in $L^1$, for example $\lVert f-f_n\rVert_{L^1}\leq n^{—1}$. Let $L(f,h):=\int_{\Bbb R}|f(x+h)+f(x)|dx$. We have \begin{align} \left|L(f,h)-L(f_n,h)\right|&\leq \int_{\Bbb R}|f(x+h)-f_n(x+h)+f(x)-f_n(x)|dx\\ &\leq \int_{\Bbb R}(|f(x+h)-f_n(x+h)|+|f(x)-f_n(x)|)dx\\ &\leq 2n^{-1}, \end{align} and we deduce that $$|L(f,h)-2\lVert f\rVert_{L^1}|\leq 4n^{-1}+|L(f_n,h)-2\lVert f_n\rVert_{L^1}|.$$ We have for each integer $n$, $$\limsup_{h\to +\infty}|L(f,h)-2\lVert f\rVert_{L^1}|\leq 4n^{—1}.$$ This gives the wanted result.
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Hint: if $f \ne 0 $ for $ x \in (a,b)$, and $h$ is large enough, $|f(x+h)+f(x)| \ne 0$ for $x \in (a,b) \cup (a-h,b-h)$.

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